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Maximize :P = 4x, + 3x2 + 2x3 3x, + 2x, + 5x, + s, = 23 2x, + x2 + X3 +Sz = 8 X, + X2 +2x, + S3 = 7 |- 4x, – 3x, - 2x3 + P = 0 %3D 3x, + 2x, +5x, 5 23 2х, + х, + x, s8 System: ST : X, + X2 +2x3 <7 х, X2, х, 2 0

Maximize :P = 4x, + 3x2 + 2x3 3x, + 2x, + 5x, + s, = 23 2x, + x2 + X3 +Sz = 8 X, + X2 +2x, + S3 = 7 |- 4x, – 3x, - 2x3 + P = 0 %3D 3x, + 2x, +5x, 5 23 2х, + х, + x, s8 System: ST : X, + X2 +2x3 <7 х, X2, х, 2 0

Practical Management Science
Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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Simplex Method Example Maximize :P = 4x, + 3x, +2x3 |3x, + 2х, + 5х, < 23 2x, + x, + x, <8 X, + x, + 2x, s7 3x, +2x, +5x, +s, = 23 2x, + x, + X3 +S2 = 8 x, + X, +2x, +s, =7 (- 4x, – 3x, -2x, +P = 0 System: ST : X1, X2, X3 2 0 X2 X3 s, S2 s, P 1 0 0 0| 23 0 1 0 0 23 +3 = 7.6 8+2=(4) 3 2 5 s2 [2] 1 1 8 1 S3 1 2 0 0 1 0 7 7÷1= 7 Р(-4) -3 -2 0 о 01 - 3R, + R, → R, 1 Row Operations:R, → R, - R, + R, → R, and 4R, + R4 → R4 X2 X3 S2 S3 P [o 1/2 7/2 1 -3/2 0 0|11]11+(1/2) = 22 1/2 1/2 0 o 0| 4 4+ (1/2) = 8 X|1 s3 0 {1/2} 3/2 0 -1/2 1 0 3 3-(1/2) = 6 P0 (-1) o 0 X1 enters to become a basic variable, s2 exits to become a nonbasic variable (pivot column identifies the entering variable and pivot row identifies the exiting variable) 1/2 2 0 1 16 R, + R¸ → R, Row Operations: 2R, → R, and R3 + R2 → R, R, + R, → R4 S3 P 1 -1 -1 0 -1 0 X, X2 X3 S, S2 S0 0 Final Tableau x, 1 0 -1 0 2 8 1 1 2 0 6 2 1 22 X2 0 1 3 0 -1 PO 0 3 0 1 X2 enters to become a basic variable, s3 exits to become a nonbasic variable (pivot column identifies the entering variable and pivot row identifies the exiting variable) s2,53 and x3 are nonbasic {=0}; So x1,X2, S1 and P are basic. Optimal Solution: Maximum P=22 at x1=1, x2=6, x3=0 (s1=8,s2=0,s3=0) Teacher's Note: Notice that x3 never entered and thus is 0. This is a viable solution. Suppose x1,x2,x3 represented 3-speed, 5-speed, 10-speed bicycles. In this problem it is more profitable not to manufacture 10-speed bikes.
Transcribed Image Text:Simplex Method Example Maximize :P = 4x, + 3x, +2x3 |3x, + 2х, + 5х, < 23 2x, + x, + x, <8 X, + x, + 2x, s7 3x, +2x, +5x, +s, = 23 2x, + x, + X3 +S2 = 8 x, + X, +2x, +s, =7 (- 4x, – 3x, -2x, +P = 0 System: ST : X1, X2, X3 2 0 X2 X3 s, S2 s, P 1 0 0 0| 23 0 1 0 0 23 +3 = 7.6 8+2=(4) 3 2 5 s2 [2] 1 1 8 1 S3 1 2 0 0 1 0 7 7÷1= 7 Р(-4) -3 -2 0 о 01 - 3R, + R, → R, 1 Row Operations:R, → R, - R, + R, → R, and 4R, + R4 → R4 X2 X3 S2 S3 P [o 1/2 7/2 1 -3/2 0 0|11]11+(1/2) = 22 1/2 1/2 0 o 0| 4 4+ (1/2) = 8 X|1 s3 0 {1/2} 3/2 0 -1/2 1 0 3 3-(1/2) = 6 P0 (-1) o 0 X1 enters to become a basic variable, s2 exits to become a nonbasic variable (pivot column identifies the entering variable and pivot row identifies the exiting variable) 1/2 2 0 1 16 R, + R¸ → R, Row Operations: 2R, → R, and R3 + R2 → R, R, + R, → R4 S3 P 1 -1 -1 0 -1 0 X, X2 X3 S, S2 S0 0 Final Tableau x, 1 0 -1 0 2 8 1 1 2 0 6 2 1 22 X2 0 1 3 0 -1 PO 0 3 0 1 X2 enters to become a basic variable, s3 exits to become a nonbasic variable (pivot column identifies the entering variable and pivot row identifies the exiting variable) s2,53 and x3 are nonbasic {=0}; So x1,X2, S1 and P are basic. Optimal Solution: Maximum P=22 at x1=1, x2=6, x3=0 (s1=8,s2=0,s3=0) Teacher's Note: Notice that x3 never entered and thus is 0. This is a viable solution. Suppose x1,x2,x3 represented 3-speed, 5-speed, 10-speed bicycles. In this problem it is more profitable not to manufacture 10-speed bikes.
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